The Fascinating World of Bravais Lattices

[aaaa202]

My book has an awful lot of text about these special lattices where everything looks the same from every lattice point. But why are these lattices so important? I mean, surely in some crystals the atomic arrangement must be such that the crystal lattice is not a bravais lattice? Edit: maybe you could also point me to a derivation of the 14 different bravais lattices as I can't really see intuitively why these exhaust all space symmetry possibilities.

 

[M Quack]

The only crystals that do not form in one of the 14 Bravais lattices are Quasicrystals, which are non-periodic.

All other crystals, from simple rock salt and silicon to protein crystals with huge unit cells for in Bravais lattices with repeating unit cells. Real crystals are of course finite and have imperfections, but the important properties are derived from the periodicity and the unit cell - band structure (conductivity, semi-conducting behaviour, etc), magnetism, birefringence, ... everything. That's why it is a big deal.

Ashcroft and Mermin has a pretty good discussion of crystal lattices.

 

[aaaa202]

But I mean, can't you draw a periodic lattice without it being a bravais lattice?

 

[M Quack]

No, you cannot. That is the whole point.

 

[aaaa202]

What about the honeycomb lattice? It is clearly periodic in some way, but it is not a bravais lattice. What exactly do you mean by periodic?

 

[M Quack]

The honeycomb lattice is a triangular Bravais lattice with a two-point basis.

http://www-personal.umich.edu/~sunkai/teaching/Winter_2013/honeycomb.html

 

[aaaa202]

So by two-point basis you mean it is a bravais lattice where we don't put an atom in the middel of the hexagons.

 

[DrDu]

So by two-point basis you mean it is a bravais lattice where we don't put an atom in the middel of the hexagons.

Exactly, a crystal consists of a basis and a lattice. The lattice is defined only by all the vectors which correspond to translations of the crystal which leave the crystal invariant.

I think there are many books on group theory who show how and why there can only be 14 lattices.

 

[LagrangeEuler]

Can you give one of them. For me is very interesting to see point groups. There is 32 point groups. When I see rotation of 1,2,3,4,6 order. Is this 5/32 point groups?

And what is liquid crystal? Do they form some crystal lattice?

 

[M Quack]

Liquid crystals are not crystals in the sense that they do not have a regular periodic lattice. In most cases they consist of long molecules that have orientational order (i.e. they point into a defined direction), but no translational order.

http://en.wikipedia.org/wiki/Liquid_crystal

The five groups you mention are all crystallographic point groups. Trivially, you also have to consider the group that contains only the identity. Additional symmetries that can be added to form the remaining groups are space inversion and 2-fold axes perpendicular to the "main" axis. For the cubic point groups, you combine 3-fold axis about the body diagonals of the cube with 2- or 4-fold axes about the faces, and sometimes 2-fold axes about the face diagonal and/or space inversion.

http://en.wikipedia.org/wiki/Crystallographic_point_group

 

[LagrangeEuler]

Thanks. One more question. What is reflection group?

 

[M Quack]

Sorry, no clue. The Wiki page is not all that helpful...

 

[sghan]

On a related note, the FCC lattice is listed as one of the 14 Bravais lattices yet it seems to really be a case of SC with a 4-atom basis. Same for BCC: SC with 2-atom basis. Can someone explain why FCC and BC are considered Bravais when they clearly reduce to a simpler lattice, while honeycomb, which also reduces, is not Bravais?

 

[DrDu]

To span a lattice, you must be able to generate the whole lattice using integer combinations of only 3 (in 2-dimensions 2) vectors. For a honeycomb structure, this is not possible. You need a basis with at least 2 atoms to generate a honeycomb lattice.

 

[sghan]

And likewise, for FCC, you need a basis with 4 atoms (for BCC 2 atoms) in order span the lattice. Integer combinations of the FCC lattice vectors do not generate an FCC lattice. Yet FCC is Bravais, why?

 

[DrDu]

No you don't, both the FCC and the BCC can be spanned with a primitive basis containing only 1 atom.
In case of the BCC:
https://commons.wikimedia.org/wiki/File:Cubic_cI_and_primitive_cell.png
and the FCC:
http://i.stack.imgur.com/TGa4T.png

 

[sghan]

Thanks DrDu! Does this mean, however, that BCC is simply a rhombohedral Bravais lattice? I do not see how BCC differs from the rhombohedral lattice.

 

[M Quack]

The reason that cubic unit cells are used for BCC and FCC is that this way the (cubic) symmetry is completely obvious. If you use the primitive unit cells that is not the case. The same argument holds for other Bravais lattices that are non-primitive, like body-centered tetragonal.

 

[DrDu]

The base centered unit cell is certainly not a rhombohedron, as the three angles aren't equal. The FCC unit cell is rhombohedral. But unlike a general rhombohedral cell, where the angles can take on any value, the angle in the FCC unit cell is fixed. So the FCC is a special case of the rhombohedral lattice with a higher symmetry. The same holds for the primitive cubic lattice which is also a special case of the rhombohedral lattice with all angles equal 90 degrees.